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13 Dec 2019

Arrange the species shown below in order of increasing bond length.

O2+ ,O2 ,O2- ,O22-

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Jamar Ferry
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17 Dec 2019

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Related questions

Arrange the species O2+, O2, O2-, O2^(2-) in order of increasingO-O bond length?

Thanks!

10.15. Which of the following species are paramagnetic: B2, N2, N22-, O2, O22-, F2?

Explain the following: (a) The peroxide ion, O22- , has a longer bond length than the superoxide ion, O2-. (b) The magnetic properties of B2 are consistent with the π2p MOs being lower in energy than the σ2p MO. (c) The O22 + ion has a stronger O—O bond than O2 itself.

The bond order is:
½ (number of bonding el. – number of antibonding el.) = ½ (8 – 6) = 1

Superoxide ion has 13 valence electrons, and one electron less in antibonding orbitals. Electron configuration of O2- ion is

The bond order is:
½ (number of bonding el. – number of antibonding el.) = ½ (8 – 5) = 1.5

The longer bond length of peroxide ion is in accordance with the lower bond order.

b) B2 molecule has 6 valence electrons. Two possible electron configurations are:

1) When σ2p MO is lower in energy:

If this is the correct el. configuration, molecule would have no unpaired electrons and would be diamagnetic.

2) When π2p orbitals are lower in energy

In this case, two energetically equivalent π2p orbitals are filled with one electron, according to Hund’s rule. As a result, the molecule is a paramagnetic with two unpaired electrons.

c) O22+ ion has 10 valence electrons, and the el. configuration:

The bond order of this ion is 3.

The bond order of O2 molecule is 2, so the O–O bond in O22+ ion is stronger.

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